Asked by kelly
                can somebody help me out? am I doing this right? using the elimination method. 
y = 3x + 2
2y-6x=4
We have the solution for y so we use this to solve
2 (3x + 2) – 6x = 4
6x + 4 – 6x = 4
0x +4 = 4
Subtract 4 from both sides
X=0
Now we use 0 for the x value in equation one
Y=3(0) +2
Y=0+2
Y=2
Points (0,2)
            
            
        y = 3x + 2
2y-6x=4
We have the solution for y so we use this to solve
2 (3x + 2) – 6x = 4
6x + 4 – 6x = 4
0x +4 = 4
Subtract 4 from both sides
X=0
Now we use 0 for the x value in equation one
Y=3(0) +2
Y=0+2
Y=2
Points (0,2)
Answers
                    Answered by
            MathMate
            
    y = 3x + 2....(1)
2y-6x=4 ....(2)
If you multiply (1) by 2 and transpose terms, you get equation (2).
This means that there is an infinite sets of solutions based on any value of t:
x=t
y=3t+2
Geometrically the system represents two lines that are coincident, hence an infinite set of solutions.
    
2y-6x=4 ....(2)
If you multiply (1) by 2 and transpose terms, you get equation (2).
This means that there is an infinite sets of solutions based on any value of t:
x=t
y=3t+2
Geometrically the system represents two lines that are coincident, hence an infinite set of solutions.
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